De Sitter space

Example

Let $n \ge 3$ and $\alpha > 0$. Equip $\R^{n+1}$ with the Minkowski form $\eta = -dx_0^2 + dx_1^2 + \cdots + dx_n^2$. The de Sitter space $dS_n$ of radius $\alpha$ is the hyperboloid $$dS_n = \set{x \in \R^{n+1} : -x_0^2 + x_1^2 + \cdots + x_n^2 = \alpha^2}$$ with the metric $g$ induced by $\eta$.

Check that it is a Lorentzian manifold: the coordinates $x_0 = \alpha\sinh(\tau/\alpha)$, $(x_1, \ldots, x_n) = \alpha\cosh(\tau/\alpha)\,\omega$ with $\tau \in \R$ and $\omega \in S^{n-1}$ cover the whole hyperboloid. In them $$g = -d\tau^2 + \alpha^2\cosh^2(\tau/\alpha)\, d\Omega_{n-1}^2,$$ where $d\Omega_{n-1}^2$ is the round metric of $S^{n-1}$, so $g$ has one negative and $n - 1$ positive directions.

The sectional curvature is constant and equal to $1/\alpha^2$, so the Ricci tensor is $R_{ab} = \frac{n-1}{\alpha^2} g_{ab}$ and $dS_n$ is an Einstein manifold. Hence it solves the vacuum Einstein equations $G_{ab} + \Lambda g_{ab} = 0$ with positive cosmological constant $$\Lambda = \frac{(n-1)(n-2)}{2\alpha^2},$$ which is $\Lambda = 3/\alpha^2$ in dimension $4$. The linear maps of $\R^{n+1}$ preserving $\eta$ preserve the hyperboloid, so the isometry group contains $O(1, n)$, and $dS_n$ is maximally symmetric. The spatial slices $\tau = \text{const}$ are spheres that shrink to radius $\alpha$ at $\tau = 0$ and then expand exponentially.